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Ok so to define Riemann integral you need to quantify over partitions. But in first-order logic you can't do that. But if you use a first-order axiomatic theory of sets (ZF for example), then quantifying over the universe is the same as quantifying over sets. Since partitions are sets, would it mean that Riemann integration can be done in first-order logic (in ZF)?

Questions:

$\rm\color{#c00}{a)}$ Is my reasoning correct? That is, can Riemann integration be done in first-order logic and more precisely in ZF?

$\rm\color{#c00}{b)}$ If the answer to $\rm\color{#c00}{a)}$ is yes, then what are the differences between a first-order and a second-order theory of Riemann integration?

$\rm\color{#c00}{c)}$ I've read that ZF is to be a framework for all of mathematics. Does this mean that all of mathematics can be done in first-order logic?

$\rm\color{#c00}{d)}$ Related to $\rm\color{#c00}{c)}$: what if my mathematical theory is that of classes (like the class of all sets)? Then this can't be done in first-order logic? Is there an axiomatic 'set' theory that treats of classes as well as sets?

Thank you very much.

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  • Your question is like asking if a function is continuous. But you're not specifying the domain, codomain, or the topologies considered. What you should do is search the website for similar questions and read their answers, which are there, since I wrote a couple of answers to each of these questions. – Asaf Karagila Jan 16 '17 at 06:35
  • (Also the first two questions are sort of related; the third and fourth are entirely different was certainly answered several times before.) – Asaf Karagila Jan 16 '17 at 09:20
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    http://math.stackexchange.com/questions/400983/how-to-write-pi-as-a-set-in-zf might provide a good start. – Asaf Karagila Jan 16 '17 at 10:26
  • By the way, I'm not the downvoter, but I can understand why people don't like your question. However, it sounds like you have come across some amount of logic and ZF set theory, and are wondering how it could even work in practice such as for Riemann integration. Hence I hope my answer helped you. I assumed a modest background in logic, so feel free to clarify any point. – user21820 Jan 17 '17 at 05:02
  • @user21820 Actually my question originated from reading the following sentence in user Todd Trimble's answer here: he wrote "But in general there is no purely first-order construction of for example the Riemann integral (involving quantification over finer and finer meshes)." Thinking about it right now, I think he meant a FO theory similar to PA, i.e. not included in some more general theory like ZF. Your answer helped, yes. Now I realize everything's just a matter of interpretation... – NeedForHelp Jan 17 '17 at 05:25
  • @user21820 ...By the way, when you write "Almost all mathematics can be translated into ZFC, yes, but almost nobody would ever attempt literally that translation for much mathematics.", I understand that, for example, a book on real analysis won't give formal proofs of the results; it is fairly clear in general that such a translation could be done and, anyway, such a translation wouldn't be of mathematical interest. I believe formal systems are primarily useful for investigating the relationships between axioms (for example axioms being inconsistent with ZF, like NFU, or independent)... – NeedForHelp Jan 17 '17 at 05:29
  • @user21820 This being said, I have read things like "The proofs will be given in an informal way" at the beginning of some books. Do they mention this because formal proofs would be better? Because such proofs really make clear and precise the logical inferences? – NeedForHelp Jan 17 '17 at 05:31
  • Your first point is indeed right. There is a spectrum of formal systems. On one end we have those that are designed for and used by the industry to design or verify machines, hardware or software. On the other end we have those that are designed to be minimal, which makes them easier to reason about but of course makes them non-usable in practice. That's okay, as long as there is a clear link between the two. Similarly, textbooks that mention the informality of their proofs are just implying that translation to a formal system might yield a much longer and less insightful proof. – user21820 Jan 17 '17 at 05:36
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    I personally believe both kinds of justifications are good to have. The truly formal proof is necessary to have essentially absolute certainty about the validity of a proof over the chosen foundational system. But the intuitive explanation is necessary to convey the key ideas to other mathematicians. One instance of this is the use of diagrams, which are immensely helpful in some cases but never shows up in a formal proof. – user21820 Jan 17 '17 at 05:37
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    Oh by first point I meant the one about investigating the relationships between axioms. As for Todd Trimble's post, if you had included it in your question you would perhaps have gotten a better reception since people would know better the context of your question. In particular, he was not talking about Riemann integration itself but rather whether definable functions over some theory are closed under it. He states that if you take the first-order theory of an exponential field, then definable functions have definable derivatives, but not necessarily definable indefinite integrals. – user21820 Jan 17 '17 at 05:47
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    More precisely, it was about definability over some particular theory that (incompletely) captures the behaviour of reals and real functions, and not over full ZFC. Namely, start with some function defined inside a first-order theory $T$, then go outside to the meta-theory ZFC and find its derivative, and then see that its derivative is definable inside $T$ as well. This works for a large class of theories $T$ such as the theory of the exponential field, but integrals don't seem to have the same nice property except for useless $T$. But I don't know whether there is any theorem to this effect. – user21820 Jan 17 '17 at 05:55
  • Following links lead to http://math.stackexchange.com/q/474034/21820. – user21820 Jan 17 '17 at 05:56
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    @user21820 Nice. Good stuff. Thanks for your dedication! – NeedForHelp Jan 17 '17 at 06:07

1 Answers1

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$\rm\color{#c00}{a)}$ Is my reasoning correct? That is, can Riemann integration be done in first-order logic and more precisely in ZF?

Yes under suitable translation. A more precise answer would require a more precise question.

$\rm\color{#c00}{b)}$ If the answer to $\rm\color{#c00}{a)}$ is yes, then what are the differences between a first-order and a second-order theory of Riemann integration?

The second-order theory of the reals does not have a countable model under full semantics, while the first-order theory of the reals does. Presumably, any theory of Riemann integration includes a theory of the reals, so it is likewise affected. Does this have any effect in practice? Apparently not. Full semantics has no effective deductive system, so even if you want to use second-order logic you would be using some deductive system that is complete for a different semantics, such as Henkin semantics. But Henkin semantics is equivalent to first-order semantics after a suitable translation, just like ZFC set theory was designed to be first-order but reflect a second-order intended semantics. So what you end up being able to prove in a second-order system of Riemman integration is surely going to be interpretable in ZFC already, unless your system includes some second-order axiom that cannot even be interpreted in ZFC.

$\rm\color{#c00}{c)}$ I've read that ZF is to be a framework for all of mathematics. Does this mean that all of mathematics can be done in first-order logic?

Almost all mathematics can be translated into ZFC, yes, but almost nobody would ever attempt literally that translation for much mathematics. Also, a small minority does mathematics in formal systems that are incompatible with ZFC, or investigates what mathematics would be like in such systems. NFU is one, whose language is the same as ZFC but whose axioms are inconsistent with ZF. One could even argue that all mathematics can be translated into PA, via the following fact: Given any practical formal system $S$ and any claimed proof $P$ of some statement $φ$ over $S$, there is a computable (and uniform) encoding of the statement "$S$ proves $φ$" as a sentence over PA denoted by "$\square_S φ$", such that PA proves "$\square_S φ$" iff $P$ is a valid proof of $φ$ over $S$, and we can computably determine whether it is so, and we can computably obtain the proof of "$\square_S φ$" from $P$ if it is so. Does that mean that PA suffices for all mathematics? Well in the above limited sense yes, but "$\square_S φ$" is devoid of meaning unless you interpret it in some 'stronger' system than PA, and down the rabbit hole you go.

$\rm\color{#c00}{d)}$ Related to $\rm\color{#c00}{c)}$: what if my mathematical theory is that of classes (like the class of all sets)? Then this can't be done in first-order logic? Is there an axiomatic 'set' theory that treats of classes as well as sets?

Look up Morse-Kelley set theory, which is a first-order theory that can reason about definable classes over ZFC, and is naturally strictly stronger. There is also second-order ZFC, and of course higher-order ZFC. How high would you like to go?

The bottom line is that whether you can do something in some formal system that can be interpreted to mean something is only meaningful if you think the syntax and the interpretation both make sense. If you strictly stick to the language of ZFC, then technically you cannot use the integral syntax nor function application syntax (namely like "$f(x)$"). You would object that it is enough if we have a (computable) uniform translation into the language of ZFC such that any statement about 'Riemann integration' is 'provable by normal means' iff their translations are provable in ZFC. (The quotes are because the term "Riemann integration" is not precise enough.) But that is too loose and will force you to also accept that 'Riemann integration' can be done in PA (as shown above in answer to (c)). You might turn around and argue that PA is merely capturing formal systems. Indeed, but can you describe what you mean by "Riemann integration" without a formal system?

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